Understanding a coin throwing game
$begingroup$
Suppose players A and B take turns to throw a fair coin. The one that
obtain $Tail$ first wins. Suppose $A$ starts. What is the probability
that $A$ wins the game? How about the probability that $A$ wins given that he did not obtain "tails" on her
first two trials. Finally, given that $A$ lost the game, what is average
number of tosses?
approach
If $A$ starts, notice that if he obtains $T$, than the game is done and this occur with probability $1/2$ If not then for $A$ to win it gotta be in the third round so we want something like HHT and $P(HHT) = 1/8$ and simiarly, if she dont win in third round then she got a change to win in fifth round and $P(HHHHT) = frac{1}{2^5}$ and so on. Thus, we have
$$ P(A ; wins) = frac{1}{2} + frac{1}{2^3} + frac{1}{2^5} + ... = sum_{i geq 1} frac{1}{2^{2i-1} } = 2 left( sum frac{1}{4^i} right) = 2 (1/[1-(1/4)] - 1 ) = boxed{2/3} $$
Now, for the second case We know for $A$ to win he must have the string HHHHT which means that $A$ to win we want to calculate :
$$ P(HHHHT) + P(H^6 T) + P(H^8 T ) + ... = frac{2}{3} - frac{1}{2} - frac{1}{8} $$
using result from previous part. this equals $boxed{1/24}$ in clear discrepancy with answer key which gives the asnwer as $1/3$. Did I misunderstood the problem?
Finally, as for the expectation if $A$ losses. We are looking at patterns of the form $HT$, $HHHT$, $HHHHHT$ remembering that $A$ was the first to start the game. If we call $X$ to be number of tosses until game ends then we observe that $P(X=2) = P(HT)$, $P(X=4) = P(HHHT)$, ... we observe that $P(X=3)$ is not possible since $A$ is to lose the game. Thus, the expectation is
$$ E(X) = sum_{i=1}^{infty} frac{2i}{2^{2i}} = 2 sum frac{i}{4^i}$$
by using the calculus identity $sum n x^n = dfrac{x}{(1-x)^2} $, we obtain
$$ E(X) = frac{8}{9} $$
again in discrepancy with my answer key which gives the solution to be $frac{8}{3}$. What is my mistake here?
probability
asked Jan 23 at 5:17
Jimmy SabaterJimmy Sabater
2,828324
$endgroup$
add a comment |
$begingroup$
Suppose players A and B take turns to throw a fair coin. The one that
obtain $Tail$ first wins. Suppose $A$ starts. What is the probability
that $A$ wins the game? How about the probability that $A$ wins given that he did not obtain "tails" on her
first two trials. Finally, given that $A$ lost the game, what is average
number of tosses?
approach
If $A$ starts, notice that if he obtains $T$, than the game is done and this occur with probability $1/2$ If not then for $A$ to win it gotta be in the third round so we want something like HHT and $P(HHT) = 1/8$ and simiarly, if she dont win in third round then she got a change to win in fifth round and $P(HHHHT) = frac{1}{2^5}$ and so on. Thus, we have
$$ P(A ; wins) = frac{1}{2} + frac{1}{2^3} + frac{1}{2^5} + ... = sum_{i geq 1} frac{1}{2^{2i-1} } = 2 left( sum frac{1}{4^i} right) = 2 (1/[1-(1/4)] - 1 ) = boxed{2/3} $$
Now, for the second case We know for $A$ to win he must have the string HHHHT which means that $A$ to win we want to calculate :
$$ P(HHHHT) + P(H^6 T) + P(H^8 T ) + ... = frac{2}{3} - frac{1}{2} - frac{1}{8} $$
using result from previous part. this equals $boxed{1/24}$ in clear discrepancy with answer key which gives the asnwer as $1/3$. Did I misunderstood the problem?
Finally, as for the expectation if $A$ losses. We are looking at patterns of the form $HT$, $HHHT$, $HHHHHT$ remembering that $A$ was the first to start the game. If we call $X$ to be number of tosses until game ends then we observe that $P(X=2) = P(HT)$, $P(X=4) = P(HHHT)$, ... we observe that $P(X=3)$ is not possible since $A$ is to lose the game. Thus, the expectation is
$$ E(X) = sum_{i=1}^{infty} frac{2i}{2^{2i}} = 2 sum frac{i}{4^i}$$
by using the calculus identity $sum n x^n = dfrac{x}{(1-x)^2} $, we obtain
$$ E(X) = frac{8}{9} $$
again in discrepancy with my answer key which gives the solution to be $frac{8}{3}$. What is my mistake here?
probability
asked Jan 23 at 5:17
Jimmy SabaterJimmy Sabater
2,828324
$endgroup$
add a comment |
$begingroup$
Suppose players A and B take turns to throw a fair coin. The one that
obtain $Tail$ first wins. Suppose $A$ starts. What is the probability
that $A$ wins the game? How about the probability that $A$ wins given that he did not obtain "tails" on her
first two trials. Finally, given that $A$ lost the game, what is average
number of tosses?
approach
If $A$ starts, notice that if he obtains $T$, than the game is done and this occur with probability $1/2$ If not then for $A$ to win it gotta be in the third round so we want something like HHT and $P(HHT) = 1/8$ and simiarly, if she dont win in third round then she got a change to win in fifth round and $P(HHHHT) = frac{1}{2^5}$ and so on. Thus, we have
$$ P(A ; wins) = frac{1}{2} + frac{1}{2^3} + frac{1}{2^5} + ... = sum_{i geq 1} frac{1}{2^{2i-1} } = 2 left( sum frac{1}{4^i} right) = 2 (1/[1-(1/4)] - 1 ) = boxed{2/3} $$
Now, for the second case We know for $A$ to win he must have the string HHHHT which means that $A$ to win we want to calculate :
$$ P(HHHHT) + P(H^6 T) + P(H^8 T ) + ... = frac{2}{3} - frac{1}{2} - frac{1}{8} $$
using result from previous part. this equals $boxed{1/24}$ in clear discrepancy with answer key which gives the asnwer as $1/3$. Did I misunderstood the problem?
Finally, as for the expectation if $A$ losses. We are looking at patterns of the form $HT$, $HHHT$, $HHHHHT$ remembering that $A$ was the first to start the game. If we call $X$ to be number of tosses until game ends then we observe that $P(X=2) = P(HT)$, $P(X=4) = P(HHHT)$, ... we observe that $P(X=3)$ is not possible since $A$ is to lose the game. Thus, the expectation is
$$ E(X) = sum_{i=1}^{infty} frac{2i}{2^{2i}} = 2 sum frac{i}{4^i}$$
by using the calculus identity $sum n x^n = dfrac{x}{(1-x)^2} $, we obtain
$$ E(X) = frac{8}{9} $$
again in discrepancy with my answer key which gives the solution to be $frac{8}{3}$. What is my mistake here?
probability
asked Jan 23 at 5:17
Jimmy SabaterJimmy Sabater
2,828324
$endgroup$
Suppose players A and B take turns to throw a fair coin. The one that
obtain $Tail$ first wins. Suppose $A$ starts. What is the probability
that $A$ wins the game? How about the probability that $A$ wins given that he did not obtain "tails" on her
first two trials. Finally, given that $A$ lost the game, what is average
number of tosses?
approach
If $A$ starts, notice that if he obtains $T$, than the game is done and this occur with probability $1/2$ If not then for $A$ to win it gotta be in the third round so we want something like HHT and $P(HHT) = 1/8$ and simiarly, if she dont win in third round then she got a change to win in fifth round and $P(HHHHT) = frac{1}{2^5}$ and so on. Thus, we have
$$ P(A ; wins) = frac{1}{2} + frac{1}{2^3} + frac{1}{2^5} + ... = sum_{i geq 1} frac{1}{2^{2i-1} } = 2 left( sum frac{1}{4^i} right) = 2 (1/[1-(1/4)] - 1 ) = boxed{2/3} $$
Now, for the second case We know for $A$ to win he must have the string HHHHT which means that $A$ to win we want to calculate :
$$ P(HHHHT) + P(H^6 T) + P(H^8 T ) + ... = frac{2}{3} - frac{1}{2} - frac{1}{8} $$
using result from previous part. this equals $boxed{1/24}$ in clear discrepancy with answer key which gives the asnwer as $1/3$. Did I misunderstood the problem?
Finally, as for the expectation if $A$ losses. We are looking at patterns of the form $HT$, $HHHT$, $HHHHHT$ remembering that $A$ was the first to start the game. If we call $X$ to be number of tosses until game ends then we observe that $P(X=2) = P(HT)$, $P(X=4) = P(HHHT)$, ... we observe that $P(X=3)$ is not possible since $A$ is to lose the game. Thus, the expectation is
$$ E(X) = sum_{i=1}^{infty} frac{2i}{2^{2i}} = 2 sum frac{i}{4^i}$$
by using the calculus identity $sum n x^n = dfrac{x}{(1-x)^2} $, we obtain
$$ E(X) = frac{8}{9} $$
again in discrepancy with my answer key which gives the solution to be $frac{8}{3}$. What is my mistake here?
probability
probability
asked Jan 23 at 5:17
Jimmy SabaterJimmy Sabater
2,828324
asked Jan 23 at 5:17
Jimmy SabaterJimmy Sabater
2,828324
asked Jan 23 at 5:17
Jimmy SabaterJimmy Sabater
2,828324
asked Jan 23 at 5:17
Jimmy SabaterJimmy Sabater
2,828324
asked Jan 23 at 5:17
Jimmy SabaterJimmy Sabater
2,828324
2,828324
add a comment |
add a comment |
1 Answer
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$begingroup$
Your answers are almost correct. The only thing you are missing are factors that account for the "given that" parts of both questions. In the first part, you are not accounting for the fact that the probability that A did not obtain tails on her first two trials is 1/8, which you need to divide by as you are conditioning on this event (it is now your sample space). You made the same error in the second problem, you need to divide by the probability that A loses, (1/3), and now your answers will match your answer key.
answered Jan 23 at 5:31
AneeshAneesh
595212
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Your answers are almost correct. The only thing you are missing are factors that account for the "given that" parts of both questions. In the first part, you are not accounting for the fact that the probability that A did not obtain tails on her first two trials is 1/8, which you need to divide by as you are conditioning on this event (it is now your sample space). You made the same error in the second problem, you need to divide by the probability that A loses, (1/3), and now your answers will match your answer key.
answered Jan 23 at 5:31
AneeshAneesh
595212
$endgroup$
add a comment |
$begingroup$
Your answers are almost correct. The only thing you are missing are factors that account for the "given that" parts of both questions. In the first part, you are not accounting for the fact that the probability that A did not obtain tails on her first two trials is 1/8, which you need to divide by as you are conditioning on this event (it is now your sample space). You made the same error in the second problem, you need to divide by the probability that A loses, (1/3), and now your answers will match your answer key.
answered Jan 23 at 5:31
AneeshAneesh
595212
$endgroup$
add a comment |
$begingroup$
Your answers are almost correct. The only thing you are missing are factors that account for the "given that" parts of both questions. In the first part, you are not accounting for the fact that the probability that A did not obtain tails on her first two trials is 1/8, which you need to divide by as you are conditioning on this event (it is now your sample space). You made the same error in the second problem, you need to divide by the probability that A loses, (1/3), and now your answers will match your answer key.
answered Jan 23 at 5:31
AneeshAneesh
595212
$endgroup$
Your answers are almost correct. The only thing you are missing are factors that account for the "given that" parts of both questions. In the first part, you are not accounting for the fact that the probability that A did not obtain tails on her first two trials is 1/8, which you need to divide by as you are conditioning on this event (it is now your sample space). You made the same error in the second problem, you need to divide by the probability that A loses, (1/3), and now your answers will match your answer key.
answered Jan 23 at 5:31
AneeshAneesh
595212
answered Jan 23 at 5:31
AneeshAneesh
595212
answered Jan 23 at 5:31
AneeshAneesh
595212
answered Jan 23 at 5:31
AneeshAneesh
595212
595212
add a comment |
add a comment |
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